CBSE 2025 · Region 7 · Set 3 · Q23 · 3 marks
An electric field $\displaystyle \overrightarrow{\mathrm{E}}$ given by: \[\begin{aligned} \vec{\mathrm{E}} & =100 \hat{i} \frac{N}{\mathrm{C}} & & \text { for } \mathrm{x}>0 \\ & =-100 \hat{i} \frac{N}{\mathrm{C}} & & \text { for } \mathrm{x}<0 \end{aligned} \] exists in a region. A right circular cylinder of length $\displaystyle 10$ cm and radius $\displaystyle 2$ cm, is placed in the region such that its axis coincides with x -axis and its two faces are at $\displaystyle \mathrm{x}=-5 \mathrm{~cm}$ and $\displaystyle \mathrm{x}=5 \mathrm{~cm}$. Calculate:(a)the net outward flux through the cylinder, and(b)the net charge inside the cylinder.
An electric field $\displaystyle \overrightarrow{\mathrm{E}}$ given by: \[\begin{aligned} \vec{\mathrm{E}} & =100 \hat{i} \frac{N}{\mathrm{C}} & & \text { for } \mathrm{x}>0 \\ & =-100 \hat{i} \frac{N}{\mathrm{C}} & & \text { for } \mathrm{x}<0 \end{aligned} \] exists in a region. A right circular cylinder of length $\displaystyle 10$ cm and radius $\displaystyle 2$ cm, is placed in the region such that its axis coincides with x -axis and its two faces are at $\displaystyle \mathrm{x}=-5 \mathrm{~cm}$ and $\displaystyle \mathrm{x}=5 \mathrm{~cm}$. Calculate:
(a)
the net outward flux through the cylinder, and
(b)
the net charge inside the cylinder.
Marking-scheme solution
\[\begin{aligned}
\phi_{L} & =\vec{\mathrm{E}} \cdot \overline{\Delta S} \\
& =100(-\hat{i}) \cdot \Delta S(-\hat{i}) \\
& =100 \times \pi\left(2 \times 10^{-2}\right)^{2} \\
& =4 \pi \times 10^{-2} \mathrm{Nm}^{2} / \mathrm{C}
\end{aligned}
\]
\[\begin{aligned}
\phi_{R} & =100(\hat{i}) \cdot \Delta S(\hat{i}) \\
& =100 \times \pi\left(2 \times 10^{-2}\right)^{2} \\
& =4 \pi \times 10^{-2} \mathrm{Nm}^{2} / \mathrm{C}
\end{aligned}
\]
\[\begin{aligned}
\phi_{\text {total }} & =\phi_{L}+\phi_{R} \\
& =8 \pi \times 10^{-2} \mathrm{Nm}^{2} / \mathrm{C} \\
& =25.12 \times 10^{-2} \mathrm{Nm}^{2} / \mathrm{C}
\end{aligned}
\]
\[\begin{aligned}
\text { Charge } q & =\varepsilon_{0} \phi_{\text {total }} \\
& =25.12 \times 10^{-2} \times 8.85 \times 10^{-12} \\
& =0.22 \times 10^{-11} \mathrm{C}
\end{aligned}
\]
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CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.